Answer: The possible values are the positive integers that are not square-free.

First, note that if we take $a=\ell$, $b=k \ell$, $c=k \ell$, $d=k^{2} \ell$ for some positive integers $k$ and $\ell$, then we have
$$\frac{a b}{a+b}+\frac{c d}{c+d}=\frac{k \ell^{2}}{\ell+k \ell}+\frac{k^{3} \ell^{2}}{k \ell+k^{2} \ell}=\frac{k \ell}{k+1}+\frac{k^{2} \ell}{k+1}=k \ell$$
and
$$\frac{(a+b)(c+d)}{a+b+c+d}=\frac{(\ell+k \ell)\left(k \ell+k^{2} \ell\right)}{\ell+k \ell+k \ell+k^{2} \ell}=\frac{k(k+1)^{2} \ell^{2}}{\ell(k+1)^{2}}=k \ell,$$
so that
$$\frac{a b}{a+b}+\frac{c d}{c+d}=k \ell=\frac{(a+b)(c+d)}{a+b+c+d}$$
This means that $a+b+c+d=\ell\left(1+2 k+k^{2}\right)=\ell(k+1)^{2}$ can be attained. We conclude that all non-square-free positive integers can be attained.

Now, we will show that if
$$\frac{a b}{a+b}+\frac{c d}{c+d}=\frac{(a+b)(c+d)}{a+b+c+d}$$
then $a+b+c+d$ is not square-free. We argue by contradiction. Suppose that $a+b+c+d$ is square-free, and note that after multiplying by $(a+b)(c+d)(a+b+c+d)$, we obtain
$$\begin{equation*} (a b(c+d)+c d(a+b))(a+b+c+d)=(a+b)^{2}(c+d)^{2} . \tag{1} \end{equation*}$$
A prime factor of $a+b+c+d$ must divide $a+b$ or $c+d$, and therefore divides both $a+b$ and $c+d$. Because $a+b+c+d$ is square-free, the fact that every prime factor of $a+b+c+d$ divides $a+b$ implies that $a+b+c+d$ itself divides $a+b$. Because $a+b<a+b+c+d$, this is impossible. So $a+b+c+d$ cannot be square-free.